Investigating the combustion of alcohols

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Introduction

Investigating the combustion of alcohols Introduction: Combustion is a process of rapid oxidation of a substance with simultaneous evolution of heat and, usually, light. Alcohol is a term applied to a group of chemical compounds that contain the OH group (oxygen and hydrogen). Alcohols have one, two or three hydroxyl, -OH, groups attached to their molecules and are thus classified as monohydric, dihydric, or trihydric, respectively. Methanol and ethanol are monohydric alcohols. Alcohols are further classified as primary, secondary, or tertiary, according to whether one, two, or three other carbon atoms are bound to the carbon atom to which the hydroxyl group is bound. Alcohols are characterized by many common reactions, the most important of which is the reaction with acids to form substances called esters, which are comparative to inorganic salts. Alcohols are normal by-products of digestion and chemical processes within cells, and are found in the tissues and fluids of animals and plants. Methanol is the simplest of all alcohols. It was formerly made by the destructive distillation of wood, however, almost all methanol produced today is synthetic, made from carbon monoxide and hydrogen. Methanol is extremely poisonous and has a relative density of 0.7915 at 20�C. Ethanol is a colourless liquid with a burning taste and characteristic, agreeable odour. ...read more.

Middle

The end temperature was recorded. This was done for all three of the alcohols twice. Diagram: Thermometer Aluminium can Clamp 100cm� H2O Flame Spirit burner Results: Results Experiment First Experiment Second Experiment Alcohol Methanol Ethanol Propanol Methanol Ethanol Propanol Water Temperature before (�C) 20 24 21 19 21 20 Water Temperature after (�C) 79 82 82 82 83 82 Mass before (g) 176.457 169.818 178.471 171.101 167.186 176.471 Mass after (g) 174.008 167.320 176.521 168.502 164.521 174.066 Temperature change (�C) 59 58 61 63 62 62 Mass change (g) 2.449 2.498 1.950 2.599 2.665 2.405 Calculations: The first of the calculations is to work out the heat given off during the combustion: Heat = Mass of water * Rise in temperature The second is to work out the number of moles of the liquid burned in the reaction: Moles = Mass/Relative Formula Mass The third of these equations uses the first two to work out the number of joules of energy given out per mole of each alcohol: Joules per mole = Energy/Moles The last equation shows the joules given out per gramme of alcohol burned: Joules per gramme = Energy/Mass change Calculations - 1st Attempt Calculation Energy Moles J/mol J/g Methanol 24780 0.076531 323789.302 10118.420 Ethanol 24360 0.054304 448582.866 9751.801 Propanol 25620 0.032500 788307.692 13138.460 Calculations - 2nd Attempt Calculation Energy Moles J/mol J/g Methanol ...read more.

Conclusion

This would have affected results, so if it were to be an absolute fair test then the wicks would have to be exactly the same size. If this experiment were to be done again then it would be timed to see which alcohol is the most efficient heater. There was much heat loss from the apparatus and if it were to be totally fair then all the heat would have to be contained during the experiment. This, however, is too expensive to do at school. Overall the experiment was conducted fairly and the results were interesting to acquire. Further Thinking: If there was enough time and resources to do so I would conduct this experiment including the improvements aforementioned in the evaluation and use other alcohols in the group like Butanol and Pentanol etc. My prediction for this would stay with my findings in this experiment, that the higher the relative formula mass of an alcohol, the more energy it will produce when it is burned in air. I would also try using an insulated aluminium can to reduce heat loss. This would produce more accurate results and it would be a fairer test. If I did this experiment again then I would take the temperature every 30 seconds to produce a graph that would show the rate of heating and I would be able to compare these at a certain point during the experiment, say 50�C. ?? ?? ?? ?? Andrew Dawson 5/1/2007 ...read more.

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